Ok, i've been working on the following problem and this is what I've gotten:
Let $F$ be a field, let $n$ be a positive integer, and let $A,B \in M{nxn} (F)$ be matrices satisfying $B\ne 0$ and $AB=0$. Show that $\lvert A \rvert =0$
Since $AB=0$ we can multiply both sides by $B^{-1}$, so $(AB)B^{-1}=0B^{-1}$ which equals $A(BB^{-1})=0= AI=0=A$
Therefore $\lvert A\rvert$ =0
I think I'm done here but I feel I may have missed something.
On
Just because $B \neq 0$, you cannot assume that $B$ is invertible. So here is what you need to think about. Since $B$ is non-zero therefore there is at least on non-zero column in $B$. let us call it $\mathbf{b}_i$. Then by the condition given you get $A\mathbf{b}_i=0$. Thus the homogeneous system $AX=0$ has a non-trivial solution, hence $A$ cannot be invertible, therefore the determinant must be $0$.
Alternately: you can say that $\mathbf{b}_i$ is an eigen vector corresponding to the eigen value $0$, hence determinant is $0$.
$B$ is not necessarily invertible, yet you're using $B^{-1}$: so your proof is wrong*.
You only proved your result for invertible $B$, not in the general case $B\neq0$.
A proof that works: since $B\neq0$, there exists a vector $X$ such that $BX\neq0$. Then, since $AB=0$ we must have $A(BX)=0$, hence $BX$ is a non-nil vector in the kernel of $A$, hence the kernel of $A$ is not nil, hence $\lvert A\rvert=0$.
* with matrices: if $B\neq0$, you can't conclude that $B$ is invertible. For example: $$B=\begin{pmatrix}1&0\\0&0\end{pmatrix}.$$ We clearly have $B\neq0$, yet $B$ is not invertible.